Determine which of the following collections of subsets of $\mathbb{R}$ are bases:
a) $C_1=\lbrace (n,n+2) \subset \mathbb{R} | n\in \mathbb{Z} \rbrace$
1) $\cup_{n\in \mathbb{Z}} (n,n+2)=\mathbb{R}$
2) Let $w_1, w_2 \in C_1$, take for example, $w_1=(1,3)$ and $w_2=(2,4)$
$\forall x \in w_1\cap w_2 =(1,3) \cap (2,4)=(2,3)$
There is no $w_3 \in C_1$ s.t. $x\in w_3 \subset w_1 \cap w_2$
So, there is no any topology on $\mathbb{R}$ such that $C_1$ is a basis for this topology.
b) $C_2=\lbrace [a,b] \subset \mathbb{R} | a<b \rbrace$
Take $[0,1]$ and $ [1,2]$, $[0,1] \cap [1,2]=\lbrace 1 \rbrace$
There is no $w_3 \in C_2$ s.t. $x \in w_3 \subset \lbrace 1 \rbrace$
c) $C_3=\lbrace [a,b] \subset \mathbb{R} | a \leq b \rbrace$
d) $C_4=\lbrace (-x,x) \subset \mathbb{R} | x\in \mathbb{R} \rbrace$
1) $\mathbb{R} = \cup_{x \in \mathbb{R}} (-x,x)$
2)Let $w_1=(-a,a), w_2=(-b,b) \in C_4$, $w_1 \cap w_2 = (-a,a) \cap (-b,b)=(-c,c)$ where $c\in \lbrace a,b \rbrace$.
So, $\forall w_1, w_2 \in C_4, \forall x\in w_1 \cap w_2=(-c,c), \exists w_3=(\frac{-c}{2}, \frac{-c}{2})$ s.t. $x\in w_3 \subset w_1 \cap w_2$
Therefore, there exists a topology in $\mathbb{R}$ and $C_4$ is a basis for it.
e) $C_5= \lbrace (a,b) \cup \lbrace b+1 \rbrace \subset \mathbb{R} | a<b \rbrace$
1) \mathbb{R}=\cup ((a,b) \cup \lbrace b+1 \rbrace$.
2) Take $w_1= ((1,2) \cup \lbrace 3 \rbrace, w_2= ((2,4) \cup \lbrace 5 \rbrace \in C_5$ we have
$w_1 \cap w_2= \lbrace 3 \rbrace$
Then, There is no $w_3 \in C_5$ s.t. $\forall x\in w_1 \cap w_2= \lbrace 3 \rbrace, x\in w_3 \subset \lbrace 3 \rbrace$
Is that’s true? And what about $C_3$?
If there exist any mathematical or Language mistakes tell me please, since I don’t speak English well.
Thanks.
Your approaches are all true.
And for $C_3$, it is trivial to show that the union of elements of $C_3$ is $\mathbb{R}$.
And by the definition, $\{r\} \in C_3$ for any $r \in \mathbb{R}$.
Thus the second condition is also trivially satisfied.